Problem: The differentiable functions $x$ and $y$ are related by the following equation: $xy=4$ We are also given that $\dfrac{dy}{dt}=1$. Find $\dfrac{dx}{dt}$ when $y=0.5$.
Solution: Let's start by differentiating the equation $xy=4$ with respect to $t$. $\begin{aligned} xy&=4 \\\\ \dfrac{dx}{dt}\cdot y+x\cdot\dfrac{dy}{dt}&=0 \end{aligned}$ We are given that ${\dfrac{dy}{dt}=1}$ and we want to find $\dfrac{dx}{dt}$ when ${y=0.5}$. We also need to find the corresponding value of $ x$. To do that, we can plug ${y=0.5}$ into the relating equation: $\begin{aligned} x\cdot y&=4 \\\\ x\cdot({0.5})&=4 \\\\ x&{=8} \end{aligned}$ Let's plug ${x=8}$, ${y=0.5}$ and ${\dfrac{dy}{dt}=1}$ into the differentiated equation: $\begin{aligned} \dfrac{dx}{dt}\cdot y+ x\cdot{\dfrac{dy}{dt}}&=0 \\\\ \dfrac{dx}{dt}\cdot({0.5})+( 8)( 1)&=0 \\\\ 0.5\cdot\dfrac{dx}{dt}&=-8 \\\\ \dfrac{dx}{dt}&=-16 \end{aligned}$ In conclusion, when $y=0.5$, the value of $\dfrac{dx}{dt}$ is $-16$.